ABCDEF is a hexagon inscribed in a unit circle with AB = CD = EF = 1. Let X, Y, Z, be the midpoints of the other three sides (i.e., of BC, DE, and FA). Prove that XYZ is an equilateral triangle.

Have had a real struggle with this one - excellent. I'm sure there's a much easier approach than with vectors. Can anyone explain the vaguely Dickensian title?

With centre, O, as origin, let the vertices have position vectors a, b, c, d, e, f,

so that the vector        ZX  =  OX – OZ   =  (1/2)(b + c)  -  (1/2)(f + a)
                       
                                                            =  (1/2)(b + c – f – a)               
                                   
 |ZX|²  =  (1/4)[b² + c² + f² + a² + 2b.c – 2b.f – 2b.a – 2c.f – 2c.a + 2f.a)

Since a to f are unit vectors, and b.a = ½ (included angle of 60^o) this gives

 |ZX|²  =  (1/4)(3 + 2b.c – 2b.f – 2c.f – 2c.a + 2f.a)

By exactly similar reasoning (or cycling the symbols),

|XY|²  =  (1/4)(3 + 2d.e – 2d.b – 2e.b – 2e.c + 2b.c) and, subtracting,

|ZX|² - |XY|² = (1/2)(– b.f – c.f – c.a + f.a – d.e + d.b + e.b +e.c)

            = (1/2)[d.b + e.(c - d) - f.(b – a) – c.a - c.f + e.b]          (1)

Looking at individual terms in (1), c – d and a – b are also unit vectors
since they represent sides of the equilateral triangles, so each scalar
product is simply the cosine of the angle between the two vectors involved.

  /DOB = /COA  so  d.b = c.a

  The angle between OE and DC is the same as /COF  so  e.(c - d) = c.f  

  The angle between OF and AB is the same as /EOB  so  f.(b – a) = e.b

So all the terms in (1) cancel out showing that |ZX| = |XY|.

By exactly similar reasoning (or cycling the symbols) |XY| = |YZ|, so
the triangle XYZ is equilateral.

Posted by Harry on 2013-08-24 13:51:41